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There is a problem with my logic and I cannot seem to point out where. There's a rocket ship travelling at close-to-c speed v without any acceleration (hypothetically), and there is an observer AA with a clock A on Earth, and there's another observer on the rocket BB with a clock B and these two clocks were initially in sync when the rocket was at rest with a FoR (frame of reference) attached to Earth. Now, this rocket's moving and AA tells us that B is running slower than A by a factor of$$ \gamma = 1 /(1-v^2/c^2)^{1/2} $$ $$ t_A/t_B = \gamma $$ where v is the relative 1d- velocity between the two i.e the earth and the rocket! That would mean the time elapsed on A is greater than that on B but this will happen only in the FoR of AA? So $t_B$ in this equation must be the time on B as observed by AA? Is this correct? What do the terms mean in the equations? If the symmetry holds and BB doesn't accelerate, then BB could say that $$ t_B/t_A = \gamma $$ right? where $t_B$ and $t_A$ are the times on B and A with respect to FoR of BB?

But I was solving this problem and I took the earth FoR of A, but the prof took the rocket FoR of B? Like how will I know which FoR to solve the problem from? It'd greatly help if the terms in all the above equations were laid down neatly! DO we even need these FoRs?? Because in all the solved problems the prof is't specifying any and is using random ones! Please help!!!!

This is the question where I messed up. The first rocket bound for Alpha Centauri leaves Earth at a velocity (3/5)c. To commemorate the ten year anniversary of the launch, the nations of Earth hold a grand celebration in which they shoot a powerful laser, shaped like a peace sign, toward the ship.

  1. According to Earth clocks, how long after the launch (of the rocket) does the rocket crew first see the celebratory laser light?

This must be 25 years. My reasoning is: If v = 3c/5 10v + vt = ct where t is time taken by the light to reach the rocket from Earth as calculated from Earth.. and I solved that for t. and added 10 years to that because the time starts at the launch of the rocket!

  1. According to clocks on the rocket, how long after the launch does the rocket crew first see the celebratory laser light?

This is 20 years. Here, I say: If it takes 25 years as observed by clocks on earth for the laser to reach the rocket, what should be the corresponding time as seen on a clock on the rocket? Using the formula:

25 = $\gamma$t where $\gamma$= 5/4

solved for t!

  1. According to the rocket crew, how many years had elapsed on the rocket's clocks when the nations of Earth held the celebration? That is, based on the rocket crews' post-processing to determine when the events responsible for their observations took place, how many years have passed on the rocket's clocks when the nations of Earth hold the celebration?

For this, I did the following: 10 years on earth = T years on rocket ship where T must be lesser than 10 as observed from Earth FoR! Therefore, T= 4(10)/5 years = 8 years! But, prof says, 10 years on Earth = T years in rocket ship where T must be GREATER than 10 as observed from the rocket FoR??? Therefore, T = 10(5/4) years = 12.5 years!!

What does this question actually want?

sai
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  • How do these numbers make sense? The distance to Alpha Centaury is about 4.4 ly. With a speed of 0.6 c it takes a little over 7 years to reach it, in the Earth's frame. So after 10 years the guys are there, at their destination and they don't move much relative to the Earth. Definitely not at 3/5c. So the laser beam will reach them in about 4.4 years in the Earth's frame. And it should be approximately the same in their frame. – nasu Jan 14 '21 at 17:55

4 Answers4

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According to clock A, clock B runs slow. According to clock B, clock A runs slow. This isn't a contradiction since events that are simultaneous in AA are not simultaneous in BB.

This will all be clear if you draw a spacetime diagram.

enter image description here

Update: to be clear, given the upvotes and comments, the spacetime diagram above is not mine but is instead found at the "spacetime diagram" link just above image and is most likely also contained in the author's book "An Illustrated Guide to Relativity".

I came upon this image for the first time this afternoon and it is one of the best I've run across to aid in visualizing the symmetric time dilation due to the relativity of simultaneity.

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    This is very cute. – Stan Liou May 03 '14 at 18:45
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    This helped me understand space-time diagrams. – Christopher King May 03 '14 at 19:55
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    Not only is this cute, clear, and pedagogically useful, it's also correctly to scale — even the clocks ($v/c=0.5$, T' at 40 minutes and 52 minutes). First-rate! – rob May 03 '14 at 22:32
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    I am taking an introductory course in special relativity and haven't come across these diagrams yet! But why are the x' and ct' oriented like that? Also ct should be t right? I'm not getting anywhere! – sai May 04 '14 at 04:39
  • @chemuser I don't think there's enough room in a comment to explanain this type of diagram well. These are more commonly known as Minkowski diagrams, you might find it easier to find helpful information under that search term. – Asad Saeeduddin May 04 '14 at 11:15
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    @chemuser, multiplying the time coordinate by the constant $c$ allows both time and space to be measured with the same unit, e.g., one light-second. The orientation of the $x'$ and $ct'$ axes come directly from the Lorentz transformation. For example, the $x'$ axis is the locus of events where $t'=0$ According to the Lorentz transformation, the equation for the $x'$ axis is $ct = \frac{v}{c}x$, a line through the origin with a slope of $\frac{v}{c}$. Like wise, the equation for the $ct'$ axix is $ct = \frac{c}{v}x$, a line through the origin with a slope of $\frac{c}{v}$ – Alfred Centauri May 04 '14 at 11:23
  • @rob This diagram can't be right. The guy is going faster than the speed of light because he is below the diagonal – user1488 Sep 20 '20 at 21:27
  • @AlfredCentauri Shouldn't he be riding up the ct' axis (instead of the x' axis) or am I confused? – user1488 Sep 20 '20 at 21:30
  • @user1488 It looks to me like Stationary Person is just to the left of the $ct$ axis, and Moving Person is just to the right of the $ct’$ axis. – rob Sep 20 '20 at 21:51
  • @user1488, mark the center of each car image and then draw a line through the centers. Note that this line (the world line of the car) is qualitatively parallel to the ct' axis and thus timelike. The x' axis is not the world line of the car. – Alfred Centauri Sep 21 '20 at 15:06
  • @Alfred I see. I was confused because the wheels are on the x' axis and the car is pointing parallel to the x' axis. I'm not sure why. – user1488 Sep 22 '20 at 16:09
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Like how will I know which FoR to solve the problem from? It'd greatly help if the terms in all the above equations were laid down neatly!

Simply put, it depends on the specifics of what the question is asking. If you're interested in what the relative difference is between the time AA measures on his own clock A and the time AA measures on his friend's clock B, then you should be using the AA frame of reference. Alternatively, if you're trying to find the difference between what BB measures on his own clock B and what BB measures on his friend's clock A, you should use the BB frame of reference. In special relativity no quantity makes sense without specifying the frame of reference it is observed in.

mhodel
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Like how will I know which FoR to solve the problem from? It'd greatly help if the terms in all the above equations were laid down neatly!

chemuser,

Your question is actually similar to this one: While Space-man lives for 1 day, then how long does Earth-man live ? 1000 years or 1 second?, and therefore the answer is similar too.

As long as you stay in the classical SR situation and consider both reference frames inertial, there is no right answer to you question. Therefore, if there are no accelerations involved (and no acceleration appears in the Lorentz transform showing time dilatation), you can always claim that the other body is time dilated. Any proof showing otherwise should actually be interpreted as falsifying the very Special Relativity Theory as it currently stands.

bright magus
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This is the question where i messed up. The first rocket [ship $B$] leaves Earth [$A$] at a velocity (3/5)c. To commemorate the ten year anniversary of the launch, the nations of Earth hold a grand celebration in which they shoot a powerful laser, shaped like a peace sign, toward the ship. [...]

  1. According to the rocket crew, how many years had elapsed on the rocket's clocks when the nations of Earth held the celebration? [...] What does this question actually want?

The referential phrase "according to" is indicative of an improper statement and treatment of the problem; the negligence (refusal?, inability?) to consider explicit geometric relations of distinct participants.

Stated properly, the aim of this question 3. is surely to determine the duration of $B$, from $B$'s initial indication of being left by $A$; but until which final indication ??

There are (at least) two different interpretations of "when the celebration was held by $A$" ($A$'s indication "ten year anniversary of the launch") as a particular indication of $B$:

  • (a): consider participant $Q$ who was and remained at rest wrt. $B$ and who was passed by $A$ just as (in coincidence with observing that) $A$ indicated the "ten year anniversary of the launch". Then (later, in the subsequent analysis of the experimental setup) refer to $B$'s indication simultaneous to $Q$'s indication of being passed by $A$ and calculate $B$'s corresponding duration $\tau B[ \circ_A, \circledS Q_A ]$.
    By the well known RT method of comparison then

$$\tau B[ \circ_A, \circledS Q_A ] = \frac{1}{\sqrt{1 - \beta^2}} \times \tau A[ \circ_B, \circ_Q ] := \frac{1}{\sqrt{1 - (3/5)^2}} \times 10 \text{ years } = 12.5 \text{ years.}$$ Or

  • (b): consider participant $P$ who was and remained at rest wrt. $A$ and who was passed by $B$ such that $P$'s indication of being passed by $B$ was (later found out to be) simultaneous to $A$ indication of the "ten year anniversary of the launch". Refer accordingly to $B$'s indication of being passed by $P$ and calculate $B$'s corresponding duration $\tau B[ \circ_A, \circ_P ]$:

$$\tau B[ \circ_A, \circ_P ] = \sqrt{1 - \beta^2} \times \tau A[ \circ_B, \circledS P_B ] := \sqrt{1 - (3/5)^2} \times 10 \text{ years } = 8 \text{ years.}$$

So: does the question mean to ask about setup/evaluation (a) or (b)?
Well, arguably participant $Q$ who was considered/required in (a) as having been and remained at rest wrt. $B$ may therefore be called "a member of the rocket crew" (even though the distance $BQ$ is accordingly

$$c~\beta \times \tau B[ \circ_A, \circledS Q_A ] := c~3/5 \times 12.5 \text{ years } = 7.5 \text{ lightyears;}$$

which certainly surpasses my preconception of a "rocket").

In my humble opinion the fair way to state the question would have been to state it properly; asking explicitly about setup case (a) or (b) or whichever was actually meant.

1.According to Earth clocks, how long after the launch(of the rocket) does the rocket crew first see the celebratory laser light? [...] My reasoning is: If v = 3c/5 10v+ vt= ct where t is time taken by the light to reach the rocket from earth as calculated from earth.

The distance values$(10~\text{years } + t) \times 3/5~c = c~t$ is apparently the distance $AJ$ between $A$ and participant $J$ who was and remained at rest wrt. $A$ and who was passed by $B$ just as (in coincidence with observing that) $B$ observed $A$'s laser signal indication.

Accordingly $t$ is the duration of $A$ from $A$'s laser signal indication until $A$'s indication simultaneous to $J$'s indication of being passed by $B$ (and observing (that $B$ observed) $A$'s laser signal indication); or equally $t$ is the duration of $J$ from $J$'s indication simultaneous to $A$'s laser signal indication until $J$'s indication of being passed by $B$ (and observing (that $B$ observed) $A$'s laser signal indication);

$$t := \tau A[ \text{anniversary}, \circledS J_B ] = \tau A[ \circ_Q, \circledS J_B ] = \tau J[ \circledS A_Q, \circ_B ].$$

(But $t$ is apparently not a duration of "the rocket crew $B$"; even though the improper formulation of the question gives this impression.)

This must be 25 years.

No: $t := \frac{10~\text{years }}{(5/3) - 1} = (3/2) \times 10~\text{years } = 15~\text{years }$.

2.According to clocks on the rocket, how long after the launch does the rocket crew first see the celebratory laser light? This is 20 years. [...]

Not quite. With the result from question (1.), similarly as argued above:

$$\tau B[ \circ_A, \circ_J ] = \sqrt{1 - \beta^2} \times \tau A[ \circ_B, \circledS J_B ] := \sqrt{1 - (3/5)^2} \times 15 \text{ years } = 4/5 \times 15 \text{ years} = 12~\text{years.}$$

user12262
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