# I Understanding the output t' from the Lorentz transform

#### TomTelford

Good afternoon,

Not sure if this should be in the homework section or not but in any case...

I'm having difficulty understanding the outputs from the Lorentz transform.

Example problem.

The earth and sun are 8.3 light-minutes apart. Ignore their relative motion for this problem and assume they live in a single inertial frame, the Earth-Sun frame. Events A and B occur at t = 0 on the earth and at 2 minutes on the sun respectively. Find the time difference between the events according to an observer moving at u = 0.8c from Earth to Sun. Repeat if observer is moving in the opposite direction at u = 0.8c.

So plunking x = 498 lightseconds, v = .8c and t = 120 the answer of ∆t(observer) = -464s or -7.7mins pops out so I've got the math right. But, I don't have an intuitive sense of what that "answer" means. Is it the amount of time shown on a clock on board the ship since the ship launch at t=0, or...?

I broke down the equation into the gamma portion and the... other factor (not sure what to call it) and they are 1.666 and -278.4s respectively. The -278.4s seems to represent the meeting point between the ship and the light sphere from the event from the stationary frame but I'm not positive. I can accept the 1.66 being the stretch/skew of the axes of the worldline from t to t' assuming that is correct as well.

Any help is appreciated.

Tom.

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#### SiennaTheGr8

I broke down the equation into the gamma portion and the... other factor (not sure what to call it) and they are 1.666 and -278.4s respectively.
Not following this part.

The earth and sun are 8.3 light-minutes apart. Ignore their relative motion for this problem and assume they live in a single inertial frame, the Earth-Sun frame. Events A and B occur at t = 0 on the earth and at 2 minutes on the sun respectively. Find the time difference between the events according to an observer moving at u = 0.8c from Earth to Sun. Repeat if observer is moving in the opposite direction at u = 0.8c.

So plunking x = 498 lightseconds, v = .8c and t = 120 the answer of ∆t(observer) = -464s or -7.7mins pops out so I've got the math right.

But, I don't have an intuitive sense of what that "answer" means. Is it the amount of time shown on a clock on board the ship since the ship launch at t=0, or...?
(What launch?)

The quantity you've found is the coordinate time that passes between Events A and B as measured in the rocket frame. The coordinate time of each event is measured locally at that event, by a clock that's co-moving with the rocket. All such rocket-frame clocks are synchronized with each other according to the rocket frame, so each of them can record the coordinate time of any events it's physically present to "witness." Later they can compare readings.

What do you make of the fact that your answer for the first part of the question is negative? What answer did you get for the second part of the question?

#### PeroK

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The Lorentz Transformation assumes a common origin. By default you took the origin to be the Earth at time $t=0$ in the Earth-Sun frame. Event A.

In the other frame, therefore, the origin must also be taken to be this event. That observer will measure things in her reference frame, with event A as the origin: $x' = 0, \ t'= 0$.

In that reference frame, event B takes place at the coordinates $x', t'$ that you calculated.

In classical physics, of course, although $x'$ would be different from $x$ for event B, we would have $t'=t$.

In SR, however, the time at which events occur depends on the reference frame.

#### TomTelford

In that reference frame, event B takes place at the coordinates x′,t' that you calculated.
Even though the observer won't actually see it until much later? In other words, I, as the ship's captain, would calculate based on my frame's time and preserving c that the event took place 7.7 minutes before I left earth (event A) and that somehow the sun is (or was?) 172 light minutes further away than someone standing on Earth would calculate?

#### Arkalius

Even though the observer won't actually see it until much later? In other words, I, as the ship's captain, would calculate based on my frame's time and preserving c that the event took place 7.7 minutes before I left earth (event A) and that somehow the sun is (or was?) 172 light minutes further away than someone standing on Earth would calculate?
Keep in mind anything in the "past" in this new frame of reference is how things look in this frame of reference. If you've just left Earth at t=0 in this frame, then the spacetime coordinates of events in the past in this frame don't represent what your own past was because you didn't experience them while stationary in that frame. If you hadn't just left Earth, but instead had always been flying that direction and are only just passing Earth, then sure, the Sun would have been much further away 7.7 minutes ago for you.

#### Nugatory

Mentor
Even though the observer won't actually see it until much later?
Yes. The way you know when something happened (that is, it's $t$ coordinate using an inertial frame in which you are at rest) is to start with the time that you see it (that is, the time on your wristwatch when the light reaches your eyes) and subtract out the time it took for the light to travel to you.

#### PeroK

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Even though the observer won't actually see it until much later? In other words, I, as the ship's captain, would calculate based on my frame's time and preserving c that the event took place 7.7 minutes before I left earth (event A) and that somehow the sun is (or was?) 172 light minutes further away than someone standing on Earth would calculate?
It's important to distinguish between raw observations and measurements in a reference frame. The travel time of light signals from an event does not mean it happened at a later time, depending on how far you are from an event. For example, on Earth you may not see events but hear them. If you are standing next to a clock and it chimes the hour, then A few seconds later you hear a clock a mile away chime the hour, you don't conclude that one clock must be wrong. You must take the travel time of sound into account.

In SR it's better to think of a reference frame as a whole network of observers, all at rest relative to each other and with synchronised clocks. All events are measured locally, then the data from all observers is collated and a picture of what happened, where and when can be put together.

#### PeroK

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Even though the observer won't actually see it until much later? In other words, I, as the ship's captain, would calculate based on my frame's time and preserving c that the event took place 7.7 minutes before I left earth (event A) and that somehow the sun is (or was?) 172 light minutes further away than someone standing on Earth would calculate?
PS think about events A and B in the Earth-Sun frame. Event A took place on the Earth at $t=0$. That wouldn't be seen on the Sun until 8 minutes later. And event B took place on the Sun at $t =2$ minutes.

That doesn't mean that an observer near the Sun concludes that event B took place before event A in their reference frame.

All observers in the Earth-Sun frame, wherever they are located (And a reference frame extends infinitely in all directions) will measure the same time and place for those events. And, indeed they will agree on the time and place of all events.

All observers in a different frame, moving relative to the Earth-Sun system will likewise agree on the time and place of every event. But, their time and place coordinates will be different from observers in the Earth-Sun frame.

The difference in the coordinates between the two reference frames is given by the Lorentz Transformation.

#### TomTelford

I guess I was really trying to overthink this. I kept trying to add back the time it took for the light from the event to get back to me.

Okay, thanks guys.

I knew that time had to slow down on the ship because if it didn't it would be seen that ship velocity towards the sun plus light velocity from the event at the sun would violate the constancy of c.

Are there any good visualization tools for this? I think where I'm uncomfortable is whenever I do any calculation I'd like to have the ability to run a good approximation in my head of what the outcome should be. With this I'm having trouble seeing it in my head.

PeroK I think that "network of clocks" is making the most sense to me.

So this is also distinct from any Doppler effect of the stationary vs. moving observers (which would be the returning light issue from above) plus things like moving up or down gravity wells, etc.?

Thanks.

#### Ibix

Are there any good visualization tools for this?
Minkowski diagrams.

They're just displacement-time graphs with the time axis vertical. But you can easily plot the axes associated with other frames, and you quickly get a feel for where an event shown in one frame ends up when viewed in another.

I wrote a tool to draw your own, with buttons to create some standard thought experiments, and to animate the frame transitions. http://ibises.org.uk/Minkowski.html

#### PeroK

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@TomTelford one of the things I found useful when learning SR is to translate the problem into the reference frame in question. In this case, in the ship frame:

The Earth and The Sun are moving towards the ship at $0.8c$. The distance between them is length contracted from their rest frame: they are about 5 light minutes apart.

The Earth clock leads (In the direction of motion) so it lags behind The Sun clock: by about 6.5 mins.

Both clocks are running slow by a factor of 3/5.

So, event B happens first when the Sun clock reads 2 mins and the Earth clock reads about -4.5 mins.

About 7.5 mins later the Earth clock finally reaches $t=0$ and event A takes place.

You'd better check my mental arithmetic there, using the Lorentz Transformation or otherwise!

"Understanding the output t' from the Lorentz transform"

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