Time Dilation: Self-Consistency Problem Explained

mathman44
Messages
204
Reaction score
0
Hi all.

I'm having trouble getting an intuitive understanding for the following situation. Let frame A and frame B be moving with relative velocity v.

It's true that a clock in frame A will be time dilated with respect to a clock in frame B, but also that a clock in frame B will be time dilated with respect to a clock in frame A.

i.e. any time between two events measured in frame A must be multiplied by gamma to get the corresponding time measured between those two events in frame B, and correspondingly, that time as measured in frame B must be multiplied by gamma to get the corresponding time as measured in frame A. Obviously this would mean that

t_a = \gamma^2{t_a}

Can anyone explain this apparent contradiction?
 
Physics news on Phys.org
mathman44 said:
i.e. any time between two events measured in frame A must be multiplied by gamma to get the corresponding time measured between those two events in frame B,
In general, that's not true. You need the full Lorentz transformations to translate the time between two arbitrary events from one frame to another.

In the special case of events that take place at the same location in A, which could be measured with a single collocated clock in A, then the simple time dilation formula would apply. (Note that events that take place at the same location in A must of necessity take place at different locations in B; so it's true that ΔtB = γΔtA, but ΔtA ≠ γΔtB.)
 
Ah... yes, if the events occur at one location in space in one frame, they must be separated in space in the other frame.

Cheers!
 
Mathman, now that Doc Al has addressed the second part of your post, can you explain how that impacts the earlier part where you said:
"It's true that a clock in frame A will be time dilated with respect to a clock in frame B, but also that a clock in frame B will be time dilated with respect to a clock in frame A."

Specifically, does his answer mean that one or both of your "true" statements are not true?

It appears to me that he was focusing on just your first statement and it appears that what you call "a clock in frame A" is what he is calling "a single collocated clock in A" but I'm wondering that if you still consider this to be a true statement, where is the clock in frame B or what clock in frame B were you talking about?
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

Similar threads

Replies
16
Views
2K
Replies
88
Views
7K
Replies
54
Views
3K
Replies
36
Views
4K
Replies
11
Views
2K
Replies
45
Views
5K
Replies
18
Views
1K
Replies
10
Views
2K
Back
Top