Schwarzschild Geometry - Shouldn't Time Speed up near a Black Hole
up vote
1
down vote
favorite
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity differential-geometry topology
asked 1 hour ago
Luke Polson
275
add a comment |
up vote
1
down vote
favorite
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity differential-geometry topology
asked 1 hour ago
Luke Polson
275
What makes you think there must be a flaw?
– Buzz
1 hour ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
1 hour ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity differential-geometry topology
asked 1 hour ago
Luke Polson
275
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity differential-geometry topology
general-relativity differential-geometry topology
asked 1 hour ago
Luke Polson
275
asked 1 hour ago
Luke Polson
275
asked 1 hour ago
Luke Polson
275
asked 1 hour ago
Luke Polson
275
asked 1 hour ago
Luke Polson
275
275
What makes you think there must be a flaw?
– Buzz
1 hour ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
1 hour ago
add a comment |
What makes you think there must be a flaw?
– Buzz
1 hour ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
1 hour ago
What makes you think there must be a flaw?
– Buzz
1 hour ago
What makes you think there must be a flaw?
– Buzz
1 hour ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
1 hour ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 34 mins ago
Thorondor
56515
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
add a comment |
up vote
5
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 50 mins ago
Javier
13.9k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448763%2fschwarzschild-geometry-shouldnt-time-speed-up-near-a-black-hole%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 34 mins ago
Thorondor
56515
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
add a comment |
up vote
3
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 34 mins ago
Thorondor
56515
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 34 mins ago
Thorondor
56515
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 34 mins ago
Thorondor
56515
answered 34 mins ago
Thorondor
56515
answered 34 mins ago
Thorondor
56515
answered 34 mins ago
Thorondor
56515
56515
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
add a comment |
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
12 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
3 mins ago
add a comment |
up vote
5
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 50 mins ago
Javier
13.9k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
add a comment |
up vote
5
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 50 mins ago
Javier
13.9k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
add a comment |
up vote
5
down vote
up vote
5
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 50 mins ago
Javier
13.9k74480
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 50 mins ago
Javier
13.9k74480
answered 50 mins ago
Javier
13.9k74480
answered 50 mins ago
Javier
13.9k74480
answered 50 mins ago
Javier
13.9k74480
13.9k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
add a comment |
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
41 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
36 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
19 mins ago
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448763%2fschwarzschild-geometry-shouldnt-time-speed-up-near-a-black-hole%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What makes you think there must be a flaw?
– Buzz
1 hour ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
1 hour ago