Question about the Galilean transform in classical physics

[billllib]

Homework Statement

Shouldn't the equation be x' = x + (v')(t) instead of x' = x + (v)(t)?

Relevant Equations

x' = x + (v)(t)?

Shouldn't the equation be x' = x + (v')(t) instead of x' = x + (v)(t)?

 

[PeroK]

Homework Statement:: Shouldn't the equation be x' = x + (v')(t) instead of x' = x + (v)(t)?
Relevant Equations:: x' = x + (v)(t)?

Shouldn't the equation be x' = x + (v')(t) instead of x' = x + (v)(t)?

What are $v$ and $v'$?

 

[billllib]

speeds in 2 reference frames. For example in A frame A's speed is zero, while in B's frame A's speed can be different then zero or = 0.

 

[PeroK]

speeds in 2 reference frames. For example in A frame A's speed is zero, while in B's frame A's speed can be different then zero or = 0.

That's not a precise answer. A possible answer is:

$v$ is the speed of B, as measured in A's frame.

$v'$ is the speed of A, as measured in B's frame.

Is that what you mean?

The convention is generally that quantities in one frame are unprimed and quantities in another frame are primed.

 

[billllib]

I am slightly modifying what you wrote.

V is the speed of A, as measured in A's frame. V = 0

V′ is the speed of A, as measured in B's frame. V !=0 or V = 0

Can that also work?

 

[PeroK]

I am slightly modifying what you wrote.

V is the speed of A, as measured in A's frame. V = 0

V′ is the speed of A, as measured in B's frame. V !=0 or V = 0

Can that also work?

What do you mean by working? What equations can you write down with those values of $v$ and $v'$?

 

[billllib]

Lets say B = 100 km/h in A's frame, A = 0.
Lets say A = -100 km/h in B's frame.

Lets focus on A in both frames. Is it correct to say V_A' = -100 and V_A = 0? Is this the correct definition of prime and not prime?

This brings me back original question.
Shouldn't the equation be x' = x + (v')(t) instead of x' = x + (v)(t)?

 

[PeroK]

Lets say B = 100 km/h in A's frame, A = 0.
Lets say A = -100 km/h in B's frame.

Lets focus on A in both frames. Is it correct to say V_A' = -100 and V_A = 0? Is this the correct definition of prime and not prime?

That's all true.

This brings me back original question.
Shouldn't the equation be x' = x + (v')(t) instead of x' = x + (v)(t)?

What are $v$ and $v'$ here?

 

[PeroK]

The idea of the Galilean transformation is as follows. Let's say you are standing $100m$ from a flag, The flag is fixed at $x = 100m$ in your reference frame. Your friend begins to walk towards the flag at $1m/s$. His reference frame is moving at velocity $v = 1m/s$ in the direction towards the flag. The flag in his reference frame has position:

At $t = 0$, $x' = 100m$

At $t = 1$, $x' = 99m$

At $t = 10$, $x' = 90m$

And:

At $t = 100$, $x' = 0m$.

Now, what is the transformation between $x$ and $x'$ in that case?

 

[billllib]

That's all true.
What are $v$ and $v'$ here?

"v" and " v' " have the same value as post "A" and" A' ". A = v and A' = v'.

Sorry for sounding like a broken record.

The equation should be x' = x + (v')(t) instead of x' = x + (v)(t)?

In the title I added "shouldn't" but it should be "should".

 

[PeroK]

The equation should be x' = x + (v')(t) instead of x' = x + (v)(t)?

In the title I added "shouldn't" but it should be "should".

No. See post #9. The equation is and always will be: $x' = x - vt$, where $v$ is the velocity of the primed frame as measured in the unprimed frame.

 

[billllib]

I think I finally am getting what you are saying. I am transforming x to x'. Also since it is a constant velocity the velocity is always the same value thanks if I have any further questions I will ask.