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Shouldn't the observer on the Earth find the clock on the rocket to show less time due to time dilation and similarly in the opposite case the observer on rocket will find the clock on earth to show less time?

There is a similar question in Arthur Beiser on which there is a discussion from some time ago on Physics Forums.

Please tell if what they concluded in the end is correct or not.

Glorfindel
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Rohan Joshi
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    Please don't post cell phone photos of pages from books. (1) It's rude to cut and paste random stuff on the internet without attribution. (2) It doesn't work with search engines. (3) It doesn't work with technologies that assist blind people. –  Sep 17 '18 at 18:58
  • @BenCrowell It would be better if you could provide a link to the guidelines of PSE which dictates such restrictions--if it doesn't exist you can consider posting it on meta and provide a link here to the question. –  Sep 18 '18 at 02:57
  • Photos and attribution are pretty much uncorrelated--any text-quote can also be provided without attribution with just as much ease. I don't think it is in PSE policy to explicitly care about external search-engine results--as far as internal search-box goes, I think that is a valid concern but it doesn't automatically dictate that one just shouldn't post images. I am completely uneducated on the third point but I absolutely think it is a really valid concern, of course. @BenCrowell –  Sep 18 '18 at 03:01
  • Hi Rohan, welcome to Physics SE! Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. (@DvijMankad the first link leads to the well-received meta post about the subject, if you'd like to use it) –  Sep 22 '18 at 10:23

3 Answers3

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I think the book has some issues.

  1. If $\Delta t$ is 30 minutes in the moving frame where $v=2\times 10^8$, then $\Delta t\prime$ in the stationary frame, where $v=0$, is 40.25 minutes so that watch should say 10:40:15... plus...

  2. Travelling 30 minutes at $2/3c$, the rocket will travel some distance and thus light (i.e. the displayed time on the watch) will take some time to travel from one observer to the other.

  3. Reciprocal does not mean it is the same in all reference frames. The proper time in each frame needs to be calculated in relation to the proper time in the other frame. When 30 minutes have elapsed for the stationary observer, 30 minutes will not yet have elapsed for the moving observer.

Mick
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  • I couldnt care less about the book but there is this one thing which is nagging me.Please tell me if my approach is flawed: According to special theory of relativity, the observer on earth would observe that for the moving observer time is dilated,i.e., the fundamental unit for time say "second" will feel normal to the observer at rest but he will feel that the "second" in the frame of the observer is longer than in his frame.That leads to him observing that a clock in the moving observers frame is showing less time elapsed in comparison to his own clock.Am i thinking right? – Rohan Joshi Sep 17 '18 at 09:10
  • Yes, you are thinking right. There is a common question like this that you can probably find by googling which uses the example of a rocket going to Alpha Centauri at close to the speed of light, and an Earth observer 'sees' that it takes about 4 years but the astronauts 'see' that it takes a very much shorter time. – Mick Sep 17 '18 at 10:48
  • The question confuses the issue by mentioning a telescope (which makes a huge difference to the usual calculation, light travel time) but doesn't seem to use it in the answer. – m4r35n357 Sep 17 '18 at 12:41
  • -1: I can't understand what you mean in your first point--can you please elaborate? The second point is, of course, correct. The third point is not correct. The symmetry argument used in the book is actually correct. If Ram observes that $\frac{30}{\gamma}$ minutes have elapsed in Shyam's clock when his own clock shows that $30$ minutes have elapsed then Shyam will also observe that $\frac{30}{\gamma}$ minutes have elapsed in Ram's clock when his own clock shows that $30$ minutes have elapsed. There is no contradiction here because simultaneity is frame-dependent in relativity. –  Sep 17 '18 at 13:15
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The book is clearly and blatantly wrong! There are more than one issues. I would lay down the major conceptual issues and let you work out the details as is ideal in answering the homework questions.

There are two different questions that the book is miserably confusing with each other. One question is

"What time would Ram observe the time to be on Shyam's clock when (according to Ram) his own clock ticks, say, 10:30?"

and the other is

"What time would Ram see (i.e. while looking through a telescope) the time to be on Shyam's clock when (according to Ram) his own clock ticks, say, 10:30?"

The book claims to answer the latter but seems to employ the method to answer the first one (and fails to even do that correctly).

As you can expect, the first question simply means answering what would the actual time be on Shyam's clock (according to Ram) when Ram's clock says that it is 10:30 (according to Ram). Since, according to Ram, Shyam's clock is ticking slowly due to time-dilation, Ram would observe the display on the clock to be $10: X$ where $X$ is simply $\dfrac{30}{\gamma}$ ($<30$ as you expected! The book, sadly, messed up here in calculating $X$ to be $30\gamma$).

But, actually, the book has asked the second question (even though the book itself is simply attempting at answering the first one as I already said). And this "actually looking" business means that we should account for the time it takes for the light signal to travel from the clock of Shyam to Ram. This means, that at 10:30 in his own clock, Ram would observe the signal that left Shyam's clock when Ram's clock was displaying $10: T$ such that $\dfrac{vT}{c}=30$. And these signals would carry the information about what the dial of Shyam's clock was actually displaying (according to Ram) at $10: T$ in Ram's clock. This would be $10: Y$ such that $Y=\dfrac{T}{\gamma}$ by the same logic used in solving the first question above.

Clearly, $Y<X<30$ (and the book is wrong in its answer whether it claims to have answered the first question (in which its answer should match $X$) or the second question (in which case its answer should match $Y$)).

Edit

The symmetry argument used in the book is correct. That is to say that if Ram observes Ram observes that $X<30$ minutes have elapsed in Shyam's clock when his own clock shows that $30$ minutes have elapsed then Shyam will also observe that $X<30$ minutes have elapsed in Ram's clock when his own clock shows that $30$ minutes have elapsed. There is no contradiction here because simultaneity is frame-dependent in relativity. The similar argument goes for $Y$ as well.

  • Thats exactly what i am thinking.There is a similar question in Arthur Beiser which i think is wrong.Please take a look at my approach on the comment i have left above,assuming that there is no time difference in light reaching the observers. – Rohan Joshi Sep 17 '18 at 09:41
  • @RohanJoshi What you are saying in your comment on Mike's answer is correct. –  Sep 17 '18 at 10:26
  • @RohanJoshi I have edited my answer to include the concern about whether both of them will observe the other's clock to have slowed down or not. Indeed they both will observe the other's clock to have slowed down. –  Sep 18 '18 at 03:06
  • Can you provide me a detailed solution mathematically using lorentz transformation? – Rohan Joshi Sep 18 '18 at 15:18
  • @RohanJoshi I think my answer is based entirely on Lorentz transformation. To wit, the factor $\gamma$ you can see everywhere is the famous Lorentz factor $\dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. As far as the "detailed solution" is concerned, we don't do that here on Physics StackExchange. We only provide basic conceptual resolution to the doubt that one is facing--the details are to be worked out by one themselves. Hope it helped! (-: –  Sep 18 '18 at 17:20
  • When we use the time dilation formula directly,the time interval and not the time elapsed is dilated,like the "second" i was talking about.Since the time elapsed is inverse to time interval the time elapsed is hence lower for moving observer but if we use the lorentz transformation for time coordinates,the result matches with the solution provided in the question posted.Am i right? – Rohan Joshi Sep 19 '18 at 13:11
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Less time or more time depends on how you measure it.

Moving clock measures shorter time interval, that two (at least) synchronized by Einstein's synchronization method clocks of “stationary” reference frame.

Hence, two clocks of stationary reference frame measure greater time interval, than moving one.

That discussion and the book only lead you astray. The book by Arthur Beiser “Concepts of Modern Physics” contains completely wrong statement (Chapter Relativity, p. 11, Formula 1.5)

http://phy240.ahepl.org/Concepts_of_Modern_Physics_by_Beiser.pdf

Beiser speaks about moving observer, but gives Transverse Doppler Shift formula for stationary one. He says:

Observer moving perpendicular to a line between him and the light source measures transverse frequency shift

$$\nu=\nu_0 \sqrt {1-v^2/c^2}$$

It is wrong. Since “moving” observer’s clock run slower, he sees, that clock “at rest” is ticking $\gamma$ times faster and light received at points of closest approach is $\gamma$ times blueshifed.

The correct formula is:

$$\nu= \frac {\nu_0} {\sqrt {1-v^2/c^2}}$$

Please look for transverse Doppler Effect here:

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Or Einstein’s 1905 paper, where he speaks about moving observer and gives correct formula for a frequency as measured by moving observer (chapter 7)

http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

Observer "at rest" measures that "moving" clock tick slower. "Moving" observer measures, that clock "at rest" ticks faster than his own.

The only problem is to choose one frame an to stay within it, or to decide who is "moving" and who is at "rest".

If each of them is stubborn like sheep and shouts that it is he who is "at rest", then each of them can really think that the other is slower and come up with all kinds of nonsense.

  • If both are observers have always been in their reference frames they would feel the other one to be moving,right? So both of them would observe the time of other one to be dilated.Am i correct? – Rohan Joshi Sep 17 '18 at 16:11
  • If each of them measures the "other" clock by means of two spatially separated and Einstein - synchronized clocks then yes, each of them measures the other as being slower. This way each of them "privatizes" "stationary" reference frame, which is in fact mutual. Transverse Doppler effect vividly demonstrates, that any frame is mutual and "privatization" of the "rest" frame leads to these wonders "the other guy is always slower". There are details about synchronization, since Einstein synchronization (speed of light is isotropic) is not the only possible one. –  Sep 17 '18 at 16:21
  • https://physics.stackexchange.com/questions/402992/is-my-understanding-of-time-dilation-correct/403057#403057. Very often people see no difference between "moving" and "stationary" observer. Beiser is the good example. He thinks that they are "the same" and makes mistake while is thinking about measured by "moving" observer frequency. –  Sep 17 '18 at 16:24
  • Can you tell me a good book for introductory Special Relativity? – Rohan Joshi Sep 17 '18 at 16:30
  • It seems it was a good book by Marder "An introduction to relativity". However (this is my opinion, I don't want to insist) but start with Lorentz relativity. J. Bell in his essay "How to teach relativity" recommended this approach. But, actually, you have to gather information bit by bit. It is important to understand synchronization of clocks, measuring technique by means of synchronized clocks and problem of one - way speed of light. Einstein synchronization leads to relativity of simultaneity, reciprocity of Lorentz transformations. –  Sep 17 '18 at 16:48