# Thermal expansion of a solid body

#### eli28

Hello,

I suddenly feel quite confused with the simple equation of a body length (1d case) after being exposed to a temperature difference :
L_final=(1+alpha*dT)*L_initial while dT is the temperature difference

I try to use this equation twice -
1. I want to find the length of a body while the temperature is raised.
L_after raising the temperature =(1+alpha*dT)*L_initial
2. Now I want to find its length when the temperature is lowered to the initial temperature back. I definitely know that it should get the same initial size back, but trying to use this equation leads to:
L_after cooling down=(1-alpha*dT)*L_after raising the temperature
And this doesn't give the same initial dimension.

So, i will be happy to hear an explanation about the proper use of the equation.

#### topsquark

Forum Staff
Hello,

I suddenly feel quite confused with the simple equation of a body length (1d case) after being exposed to a temperature difference :
L_final=(1+alpha*dT)*L_initial while dT is the temperature difference

I try to use this equation twice -
1. I want to find the length of a body while the temperature is raised.
L_after raising the temperature =(1+alpha*dT)*L_initial
2. Now I want to find its length when the temperature is lowered to the initial temperature back. I definitely know that it should get the same initial size back, but trying to use this equation leads to:
L_after cooling down=(1-alpha*dT)*L_after raising the temperature
And this doesn't give the same initial dimension.

So, i will be happy to hear an explanation about the proper use of the equation.
The standard definition for (alpha) gives dL = (alpha) L dT for the linear expansion equation. I'm not sure where your equation comes from.

-Dan

#### eli28

It was presented to us in a lecture

It was presented to us in a lecture.
I guess this form is a simplification to a linear relation while the temperature difference is not big.
You can find this form of equation in wikipedia as well.

I will be happy to get an answer to this confusing (probably simple) question.

Thanks,

#### topsquark

Forum Staff
It was presented to us in a lecture.
I guess this form is a simplification to a linear relation while the temperature difference is not big.
You can find this form of equation in wikipedia as well.

I will be happy to get an answer to this confusing (probably simple) question.

Thanks,
Sorry, the dT in the equation threw me off. It implied we were using the differential version. Properly speaking it should be (Delta)T.

Anyway, let's do this. I'm going to use dT = T - T(0) for clarity. We know that
L - L(0) = a (T - T(0))

That gives
L = L(0) + a(T - T(0))

To go back things get a little tricky but not too bad if you go step by step. First let's redefine the variables by setting
L' = L'(0) + a (T' - T'(0))

Now, this takes us from length L'(0) and T'(0) to a new temperature T'. If we are starting from where we left off the heating we know that L'(0) = L from the heating and T'(0) = T. (Make sure you understand where these come from. Use numbers if you need to.) Plugging these into the L' equation we get
L' = L + a (T' - T)

But finally we know that the final overall temperature is the same as the initial so T' = T. thus
L' = L + a (T - T) = L

-Dan

#### studiot

Come on physicists this is the same fallacy that trips up economists and tax accounts. Are we not better at mathematics than they are?

Sue buys a lollipop from me for $1.00 and sells it to Dan for$1.10.

Sue makes 10% profit; the absolute change is 10c.

Dan then sells the lollipop back to me for $1.00. Thus Dan makes (1.10 - 1.00)/1.10 % loss or 9.09% loss; The absolute change is still 10c. Edit Well done Dan for revising your post2 whilst I was posting my offering. Does that make up for your 10c loss? Incidentally you will find the original equation L_final=(1+alpha*dT)*L_initial derived in such standard texts as Sears and Zemansky Modern University Physics from Dan's full equation. Last edited: #### topsquark Forum Staff Sue buys a lollipop from me for$1.00 and sells it to Dan for \$1.10.
I would never buy a lollipop from my ex-wife.

-Dan