Questions about inertial mass

[Johnny5]

I already gave the equation. F.d (dot product). By definition, the force can do no work. The direction will change, but not the energy.

 

Direction will change but not energy. Hmmm.

 

The direction of the applied force is in the j^ direction.

 

The direction of the change in position due to the applied force is also in the j^ direction.

 

The force has the same direction as dR, where dR is the infintessimal change in position (along the y axis) due to an infinitessimal force (along the y axis).

 

Dot product is simple enough...

 

[math] \vec A \bullet \vec B = | \vec A | |\vec B | cos \theta [/math]

 

Where theta is the angle between vector A, and vector B.

 

So lets see...

 

The two vectors have the same direction, and I might also add the same line of action (sometimes this is important you know; specifically I mean in analysis of how an applied force rotates something instead of imparting pure translation, i can explain this to you later), so that the angle between the two vectors is zero, and the cosine of zero is 1. So where am i going with this...

 

Oh yeah... you said the applied force does no work.

 

[math] \vec F \bullet d\vec R = F dR [/math]

 

since theta is zero.

 

Work is defined to be the integral of F dot dr.

 

[math] W = \int F dr [/math]

 

Where the integral is taken from the initial R position, to the final R position. So let me write this more clearly for myself.

 

[math] W = \int_{r=r1}^{r=r2} F dr [/math]

 

I can pull F out of the integral, provided that it is a constant, so far as the integral is concerned. Here we just have someone punch the thing. So its an impulse. Taking F out of the integral we have:

 

[math] W = F \int_{r=r1}^{r=r2} dr [/math]

 

So that I get this:

 

[math] W = F (r2-r1) [/math]

The only way for W to equal zero, is either F =0 or r2=r1. F isn't zero by stipulation, and r1 and r2 are different, since the proton did move upwards slightly.

 

Where on earth are you coming up with W=0?

[swansont]

If you do the integral the way you did you aren't modeling the situation properly. If the force is applied in a constant direction for a macroscopic length of time, it will not be perpendicular for the whole time. If you stipulate that it is perpendicular then it isn't constant (it changes direction) so you can only pull the magnitude out of the integral, not the unit vector. The dot product is still there, and is still zero.

 

i.e. Saying the force is perpendicular to the displacement is not the same as saying the force is always in the j direction.

[Johnny5]

If you do the integral the way you did you aren't modeling the situation properly. If the force is applied in a constant direction for a macroscopic length of time' date=' it will not be perpendicular for the whole time. If you stipulate that it is perpendicular then it isn't constant (it changes direction) so you can only pull the magnitude out of the integral, not the unit vector. The dot product is still there, and is still zero.

 

i.e. Saying the force is perpendicular to the displacement is not the same as saying the force is always in the j direction.[/quote']

 

Thank you, I think I am slowly getting it. I went back and read this thread from the beginning. I will have to read it a few more times, until I finally am certain of what you are trying to say.

 

I am still confused on the whole "mass is a vector" idea.

 

I went back and reread from here:

 

F = Mdv/dt+vdM/dt

 

In the case where M is relativistic mass, the force formula has two nonzero terms on the RHS. Particularly, dM/dt is nonzero, if M is relativistic mass.

 

I understand this much for sure.

 

But somewhere I think you said that the formula for relativistic mass is wrong because it doesn't contain any vector symbols, so I still don't fully understand what you are trying to say. But I am trying.

 

Thank you

[Johnny5]

I'm saying that if you use that as your definition of mass, then you get a given force causing different accelerations in different directions. You have no velocity in a direction perpendicular to your motion, thus you get a different answer. Your term for mass that is direction dependent, i.e. it's no longer a scalar. That's the problem. Using rest mass and relativistic formulae alleviates this problem.

 

I think I am beginning to understand you.

 

Though you really aren't being clear enough for me.

 

Correct me if I am wrong, but i think you are saying this...

 

You are saying that if I use "relativistic mass" as my definition for mass, instead of "rest mass" as my definition for mass, then what I have defined to be mass is now a direction dependent quantity.

 

Is that what you are saying?

 

Regards

[Johnny5]

If you want to use the inertial idea that m = F/a' date=' the only way to do that is if the mass is no longer a scalar[/quote']

 

Can you explain this better, I think it's your whole point.

 

Thank you again for taking the time.

 

Regards

[swansont]

You are saying that if I use "relativistic mass" as my definition for mass' date=' instead of "rest mass" as my definition for mass, then what I have defined to be mass is now a direction dependent quantity.

 

Is that what you are saying?

[/quote']

 

Yes