Simple problem about relativity and train

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The discussion revolves around solving a relativity problem involving a train and a passenger running inside it. The initial approach attempts to calculate the time elapsed for the passenger's run using the train's frame, leading to confusion when transitioning to the ground frame. The error arises from not accounting for the Lorentz Transformation, which incorporates both time dilation and the relativity of simultaneity. To arrive at the correct answer, one must first analyze the situation from the passenger's frame before converting to the ground frame. Understanding these concepts is crucial for accurately solving problems in special relativity.
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Homework Statement
A train of proper length L moves at speed v1 with respect to the ground.
A passenger runs from the back of the train to the front at speed v2
with respect to the train. How much time does this take, as viewed by
someone on the ground?
Relevant Equations
...
I want to solve this question first using the train frame, and so going to the ground frame, but the things got wrong, so i would aprpeciate to know why.

I mean, there is another ways to solve it, but i want to know where is the error here.

Simply, in train frame the time elapsed between the event 'runs from the back' and 'coming in the front' is $$L/v2.$$

Now, why can't we go direct from this time to the ground interval? That is, $$L\gamma_{2}/(v2)$$

Actually, to get the right answer, we need to go first to person frame, and so to ground frame.
 
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Herculi said:
Homework Statement:: A train of proper length L moves at speed v1 with respect to the ground.
A passenger runs from the back of the train to the front at speed v2
with respect to the train. How much time does this take, as viewed by
someone on the ground?
Relevant Equations:: ...

I want to solve this question first using the train frame, and so going to the ground frame, but the things got wrong, so i would aprpeciate to know why.

I mean, there is another ways to solve it, but i want to know where is the error here.

Simply, in train frame the time elapsed between the event 'runs from the back' and 'coming in the front' is $$L/v2.$$

Now, why can't we go direct from this time to the ground interval? That is, $$L\gamma_{2}/(v2)$$

Actually, to get the right answer, we need to go first to person frame, and so to ground frame.
When comparing the times of two events, the train frame and the ground frame are related according to the Lorentz Transformation (not simply time dilation). There is a combination of time dilation and the relativity of simultaneity.
 
In addition to the formal Lorentz transformation approach, you could also use length and velocity observed in the ground frame and follow kinematics . Length can be found through length contraction and velocity can be found from relative velocity.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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