0
$\begingroup$

Lorentz's equations show the elapsed time between two events in moving frame is greater than rest frame. But moving observer's clock ticks slower. Isn't that an inconsistency?

Consider two observers, Earthman and Spaceman in two standard inertial frames S and S' respectively. Earthman creates two events, both at the origin of his coordinate system; the first event occurs at $t=0$; the second at $t=T$. Therefore Earthman perceives the elapsed time between the events is $T$.

Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock. Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T as shown below.

$$t'=T/\gamma$$

However, Earthman applies the following Lorentz Transformation to conclude that Spaceman's measure of the elapsed time between the two events is greater than $T$ as shown below.

$$t'= \gamma(t-xv/c^2) = \gamma(T-0v/c^2) = \gamma T$$

$\endgroup$
7
  • $\begingroup$ Presumably in the first instance you mean to refer to the twin "paradox" and in the second you are talking about plain time-dilation effects? If so the apparent inconsistency is explained by the changing planes of simultaneity associated with acceleration. Getting the two issues mixed up is a ubiquitous problem when learning the subject, and you have to get them sorted out to make progress. $\endgroup$ Commented Aug 15, 2019 at 0:32
  • $\begingroup$ @dmckee, it's interesting to me that the current VTC is for "homework-like questions..." reason. It didn't occur to me that this is a homework-like question but the more I think about, it does seem like the kind of question a devious physics GTA might come up with. $\endgroup$ Commented Aug 15, 2019 at 0:59
  • $\begingroup$ Possible duplicate of Lorentz transformation from resting to moving frame: Does it matter when the origins of two coordinate systems coincided? $\endgroup$
    – Rob
    Commented Aug 15, 2019 at 5:54
  • $\begingroup$ After some time to reflect, I've decided to delete my answer (still visible to those with reputation $\ge$ 10k), and VTC for the "off-topic, homework-like question" reason. $\endgroup$ Commented Aug 15, 2019 at 13:16
  • $\begingroup$ I may not have been clear about who perceived what. Since the elapsed time between the events that Earthman created was T, then Earthman's measurement of elapsed time between the events was T. But Earthman perceived that Spaceman's measurement of elapsed time would be less than T because his moving light-clock ticks less often than earthman's. But Earthman perceived that Spaceman's measurement of elapsed time would be greater than T because of Lorentz's equation. Thus Earthman perceived conflicts in Spaceman's measurement of elapsed time. $\endgroup$
    – Richard
    Commented Aug 15, 2019 at 21:34

3 Answers 3

2
$\begingroup$

Earthman creates two events, both at the origin of his coordinate system; the first event occurs at $t=0$; the second at $t=T$. Therefore Earthman perceives the elapsed time between the events is T

Earthman observes that the elapsed time between the events is $T$. It's crucial to note that Earthman requires a single clock at relative rest to measure this elapsed time.

Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock. Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T

Here is where your reasoning fails. It's true that Earthman observes Spaceman's clocks to tick less often. But it's also true that the clock Spaceman uses to record the time of the first event is not the clock that is used to record the time of the second event (remember, the 2nd event is at a different location then the first event - according to Spaceman - so, two clocks are needed by Spaceman, one at rest at the location of the first event, and the other at rest at the location of the second event).

So Spaceman must subtract the time reading on one clock from the time reading on another clock in order to determine the elapsed time between the events.

According to Spaceman, the two different clocks are synchronized but Earthman observes that the Spaceman clock used to record the time of the second event is ahead of the Spaceman clock used to record the time of the first event!

According to Earthman, Spaceman calculates a longer elapsed time (due to the unsynchronized clocks) even though Earthman observes the two clocks run slower than his clock.

So it just isn't true that "Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T"

Instead, Earthman observes that Spaceman has calculated the elapsed time with unsynchronized clocks that are running slower than his clock.

$\endgroup$
1
  • $\begingroup$ Of course my above statement is incorrect. I took into consideration Spaceman's relative velocity in ascertaining his light clock ticked less often, but I totally ignored that his velocity placed him at a different location in space when the second event occurred. Thanks for pointing out my error. $\endgroup$
    – Richard
    Commented Aug 16, 2019 at 4:03
0
$\begingroup$

So, you have two events, $A:(x=0, t=0)$, and $B:(x=0, t=T)$.

However, Earthman applies the following Lorentz Transformation to conclude that Spaceman's measure of the elapsed time between the two events is greater than T as shown below.

Correct. Spaceman will measure the time difference between the events, $t'$, as being greater than $T$.

Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock. Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T as shown below.

Incorrect. What "Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock" means is that when Earthman looks at two events that have the same value of $x'$, $\Delta t'$ for those two events will be lower than $\Delta t$. Remember, Earthman and Spaceman disagree not just about time, but also space. When Spaceman looks at his clock, he sees it as moving through just time. Earthman sees it as moving through space as well, and thus sees it as moving through more time.

When Earthman and Spaceman look at two events that Spaceman says happen at the same location (such as a clock that is, in Spaceman's reference frame, stationary), Earthman measures there being more time between those events than Spaceman does. When they look at an event that Earthman says happen at the same location, Earthman measures there being less time than Spaceman does.

$\endgroup$
1
  • $\begingroup$ Of course my above statement is incorrect. I took into consideration Spaceman's relative velocity in ascertaining his light clock ticked less often, but I totally ignored that his velocity placed him at a different location in space when the second event occurred. Thanks for pointing out my error. $\endgroup$
    – Richard
    Commented Aug 16, 2019 at 4:02
0
$\begingroup$

Lorentz's equations show the elapsed time between two events in moving frame is greater than rest frame. But moving observer's clock ticks slower. Isn't that an inconsistency?

I am assuming that the moving frame actually refers to the frame relative to the observer at rest, and the rest frame refers to the frame relative to the moving observer. In addition, I also believe that a better phrase than ‘elapsed time’ would be ‘perceived time’, since ‘elapsed time’ brings with it an idea that absolute time exists, when in fact it does not exist in special relativity.

I do not believe there is an inconsistency. While what you’ve said is true, that the moving observer’s clock ticks slower when observed by the observer at rest(in the moving frame), this is only true for the moving frame. Once you change frames, the perceived time would change. To the moving observer(who is in the rest frame), his clock would appear to tick normally while the stationary observer’s clock would appear to tick slower.

Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T as shown below.

𝑡′=𝑇/𝛾

I believe your equation may be wrong since the Lorentz factor 𝛾 = 1/(1+v^2/c^2). For t′ to be less than T, you should be multiplying by 𝛾 and not dividing by 𝛾. If so then there is no inconsistency with the Lorentz transformation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.